If an object is conductive what do you know it can do?

Physics

1 answer:

fredd [130]3 years ago

3 0

Explanation:

some objects that are conduct :Metals conduct electricity, so if there is just the symbol of a metal, eg Mg(s), then it will conduct, even in the solid or liquid state. 2. Ionic compounds conduct when they are dissolved in water, or are melted.

what can it do ?:

A conductor is a material which electricity, heat or sound can flow through. An electrical conductor conducts electricity. ... This means that they make it very hard for electricity to flow through them. A material that stops electric current is called an insulator (electricity).

hope this helped! if not I'll answer again

You might be interested in

Describe how switching the desk lamp on and off shows that light waves transfer energy

Ludmilka [50]

The lamp transfers electric energy into both light and thermal energy when it turns on

4 0

3 years ago

Read 2 more answers

Three dogs (Spot, Fido, and Steinberg) are pulling on a chew toy. The chew toy is experiencing no acceleration. Spot is pulling

quester [9]

Answer:

The magnitude and direction of the force applied by Steinberg are approximately 15.192 newtons and 126.704º.

Explanation:

The chew toy is at equilibrium and experimenting three forces from three distinct dogs. The Free Body Diagram depicting the system is attached below. By Newton's Laws we construct the following equations of equilibrium: (Sp is for Spot, F is for Fido and St is for Steinberg) All forces and angles are measured in newtons and sexagesimal degrees, respectively:

$\Sigma F_{x} = F_{F}\cdot \cos \theta_{F} + F_{St,x} = 0$ (1)

$\Sigma F_{y} = F_{F}\cdot \sin \theta_{F}-F_{Sp}+F_{St,y} = 0$ (2)

If we know that $F_{F} = 20\,N$ , $F_{Sp} = 30\,N$ and $\theta_{F} = 63^{\circ}$ , then the components of the force done by Steinberg on the chewing toy is:

$F_{St,x} = -F_{F}\cdot \cos \theta_{F}$

$F_{St,x} = -(20\,N)\cdot \cos 63^{\circ}$

$F_{St, x} = -9.080\,N$

$F_{St,y} = F_{Sp}-F_{F}\cdot \sin \theta_{F}$

$F_{St,y} = 30\,N-(20\,N)\cdot \sin 63^{\circ}$

$F_{St, y} = 12.180\,N$

The magnitud of the force is determined by Pythagorean Theorem:

$F_{St} = \sqrt{F_{St,x}^{2}+F_{St,y}^{2}}$

$F_{St} =\sqrt{(-9.080\,N)^{2}+(12.180\,N)^{2}}$

$F_{St} \approx 15.192\,N$

Since the direction of this force is in the 3rd Quadrant on Cartesian plane, we determine the direction of the force with respect to the eastern semiaxis:

$\theta_{St} = 180^{\circ} + \tan^{-1} \left(\frac{F_{St,y}}{F_{St,x}}\right)$